ࡱ> 02/y#` bjbj\.\. .L>D>DS '''8'l\(Ltd2(D*+++,,,ddddddd$eh2hv=d,],",,,=d++ORdw/w/w/,8++dw/,dw/w/N^c+( O^T'-`0dhd0d`$h/Lh`chc@,,w/,,,,,=d=d[/,,,d,,,,tttD'ttt'TF6| Worksheet # The least Common Multiple (LCM)___ Name___________________ The Least Common Multiple (L.C.M.) of a set of numbers is the smallest number that each of the numbers in the given set can divide into it with a remainder of zero. To add or subtract fractions with different denominators, the least common multiple of the denominators is most convenient denominator to use. One method of finding the least common multiple is to write down the multiples of the given numbers in a list. Then find what numbers are shared by the numbers in the lists, and pick the smallest. Using this method, we can define the least common multiple as the smallest number in the set of common multiples. The least common multiple of a and b is the least nonzero number that is a common multiple of a and b. Example 1 Find the common multiple of 24 and 18. Solution: Multiples of 24: { 24, 48, 72, 96, 120, 144, 168, 192, 216, 240 } Multiples of 18: { 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, } Common Multiples of 12 and 18: {72, 144, } The least common multiple of 24 and 18 is 72. Problem 1 Find the least common multiple of a) 12 and 16. b) 39 and 52 c) 24 and 96 Another method of finding the least common multiple is to first write each number as a product its prime numbers. Then multiply together all the prime factors, using the common factors only once and taking only the highest exponent of each prime factor. In this case you must express each number or expression in an exponential form and select the factors with the largest exponent without repeating any of the factors. Example 2 Find the Least Common Multiple of 45, 54, and 24. Solution: Write each number as a product of primes. 45 = (9)(5) = (3)(3)(5) = 32  EMBED Equation.3 5 54 = (18)(3) = (2)(3)(3)(3) = 2 EMBED Equation.3 33 24 = (8)(3) = (2)(2)(2)(3) = 23  EMBED Equation.3 3 The different factors appearing in all the terms are 2, 3, and 5. maximum of all the 2s, (2, 23) = 23, maximum of all the 3s, (32, 33, 3) = 33, maximum of all the 5s, (5) = 5. Therefore the LCM = 23 EMBED Equation.3  33 EMBED Equation.3 5 = (8)(27)(5) = 1080 Problem 2 Find the LCM of 45 and 60. b) Find the LCM 12, 16, and 48 Example 3 Find the Least Common Multiple of 45a2bc5 and 27ab2c3d. Solution: Write each number as a product of primes. 45 = (9)(5) = (3)(3)(5) = 32  EMBED Equation.3 5 45a2bc5 = 32  EMBED Equation.3 5 a2 b c5 we have, 27 = (9)(3) = (3)(3)(3) = 33 27ab2c3d = 33 ab2c3d The different factors appearing in all the terms are 3, 5, a, b, c, and d. maximum of all the 3s, (32, 33) = 33, maximum of all the 5s, (5) = 5, maximum of all the as, (a2, a) = a2, maximum of all the bs, (b, b2) = b2, maximum of all the cs, (c5, c3) = c5, maximum of all the ds, (d) = d. Therefore the LCM = 33 EMBED Equation.3 5 EMBED Equation.3  a2 EMBED Equation.3 b2  EMBED Equation.3 c5 EMBED Equation.3 d = (27)(5)a2b2c = 135a2b2c5d Problem 3 Find the LCM of 8p4t3 and 36p2t5r. Find the LCM of 2x2y3z, 6xy2z2, and 4x3y3z3. Find the LCM of 5(x 2)2, 2(x 2)3, and 10(x 2). Find the LCM of 12(x 1)3(y + 1)6z6 , 6(x 1)4(y + 1)4z4, and 4(x 1)2(y + 1)5z3. The Least Common Denominator (LCD) The least common denominator of several fractions or rational expressions is the LCM of the individual denominators. LCD is the smallest number that is evenly divisible by all of the denominators of the given fractions. To add or subtract fractions with different denominators, the least common multiple of the denominators is most convenient denominator to use. Example 4 Determine the LCD of the following fraction and then determine an equivalent fraction with the LCD as the new denominator.  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3  Solution: the denominators are 15x, 35y2, and 9x2y. Write each number as a product of primes. 15 = (3)(5), 35 = (5)(7), and 9 = (3)(3) = 32 The different factors appearing in all the terms are 3, 5, 7, x, and y. maximum of all the 3s, (3, 32) = 32, maximum of all the 5s, (5, 5) = 5, maximum of all the 7s, (7) = 7, maximum of all the xs, (x, x2) = x2, maximum of all the ys, (y2, y) = y2. LCD = (32)(5)(7)( x3)( y2) = . Equivalent fractions:  EMBED Equation.3  Problem 4 Determine the LCD of the following fractions a)  EMBED Equation.3 ,  EMBED Equation.3  b)  EMBED Equation.3 ,  EMBED Equation.3  c)  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3  d)  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3  e)  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3  f)  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3      2001 Michael Aryee Least Common Multiple Page  PAGE 1 The least common multiple is the smallest number in the set of common multiples. 3@EPXY= R j   U i p v w z ( ) . / f q ƷƷƪƷƕvjhCJUmHnHuhMHh5 h5\)h)<hB*CJOJQJ^JaJphhMHh6B*phhB*ph h)<hh)<hB*phh56\] h6]h hQh h>*hQh>* h.m>*.XY 1 2 = > e f q r * + Z [  !gdgdgd$a$gd$a$gd $dha$gddhgdgd / E V Z  7 d hlu  "#$%;HMRWX[jjhEHUjhEHUjhEHUj&= hCJUVaJjhU hH*hh5 hc6h h6] h=7Ghhc6h5hQh5h h5\8[ \ f g klvwh^hgd & F ^gddhgd $dha$gddhgdgdgdAzZ[EGHSgd dh^gdgd & F ^gd dh^gddh^`gdh^hgd & F ^gd ^`gd[b{|,-./@ADEGùï餜 hh5\h56\ha*h56\jhEHUjhEHUj&= hCJUVaJjhU hc6hh5H*\ h5\ hH*h h6]8STrstuvw ^`gdh^hgd & F ^gddhgd8^8gd & Fgdgd6789LMNOmnpqxyz{ &'*+,-01Taftz{~Ҿh5H*\ h6] h5\ hhj> hEHUjhEHUj&= hCJUVaJjhU hH*hhh5 h5?/0}~*V345?@fgh & Fgdgdgd & F ^gd dh`gdh^hgd & F ^gd ^`gd%&'*+2>?EFIJOPQTUV]ij}¸®¤j]hEHUj hEHUj hEHUj&= hCJUVaJjhU h6]h5H*\ h5\ hH*hc6h6h?!"#$(+,-./0125>TUVWX_`abcd{|}~涩ڐډډډډڄډڄ hH* h6]hhh5H*\]ha*h5H*\ha*h56\]ha*h56\jhEHUj&= hCJUVaJhh5H*\ h5\jhUjhEHU4hi,-./RSGH $dha$gd$a$gd  !gd8^8gd & Fgdgd  !"'()*/R/mJK^_`acdwj|h|hEHUjA hCJUVaJjhU hh5\ h)<hh)<hB*ph h1Qh h5\hc6h6 hH* h6]h>wxyz+,-S`epuvxȻ׶צסצייייג׃גג׃סג׃ג׃ג h6]h5H*\ h5\h:h5 hH*hv h6h|h5 h5jh:hEHUjwB hCJUVaJhjhUjh:hEHUjA hCJUVaJ3H./0x*XY{|gd dh`gddh^`gddhgd$a$gd%&'*,3?@HIJLMRSTVWXY\`cdeghjkmnoprstuǽjhJ"hEHUjB hCJUVaJjhUhJ"h5\h:h5H*h:h5 h5h5H*\ h5\ hH*hJ"h6 h6 h6]h567JKLMOPcdefgnors}pib h1Qh hwebhj$h`hEHUjB hCJUVaJj"h|hEHUjB hCJUVaJj h`hEHUjؼB hCJUVaJjh|hEHUjԼB hCJUVaJjhUh`h5 hh5h h:h"ghijklmno)*+,-./012$a$gd$dd[$\$a$gd $dha$gd #$Ȼ׬אtgXjB hCJUVaJjx/h:hEHUjB hCJUVaJjE-h`hEHUjB hCJUVaJj)+h`hEHUj`B hCJUVaJj(hhEHUjzB hCJUVaJhjhUj&hhEHUjpB hCJUVaJ$%&()1256IJKLNObcdeklɼ歠摄uhYj5B hCJUVaJjp:h`hEHUjTB hCJUVaJj8h:hEHUjB hCJUVaJj5h`hEHUjB hCJUVaJj3h`hEHUjB hCJUVaJ h1Qh hwebhhjhUj1h`hEHU"ORw|}û h5\h^/h["yCJmHnHujhf<CJU hf<CJhf< hf<CJh["yjh["yU hhf<j>h`hEHUjJB hCJUVaJhjhUj<h:hEHU#2gd &d P $dd[$\$a$gd,1h/ =!"#$% Dd J  C A? 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